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Finding function continuity and derivatives 
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#1
May2406, 07:28 AM

P: 43

I'm not sure where to start with this question. If a limit was given, I could solve it but without it given, I am completely lost...
State on which intervals the function [tex]f[/tex] defined by [tex]f(x) = \left\{\begin{array}{cc}x + 1,&x < 0\\x^2 + 1,&x \geq 0\end{array}\right.[/tex] is: i) continuous ii) differentiable Find the derivative [tex]f'(x)[/tex] at all points where the function is differentiable. 


#2
May2406, 08:05 AM

P: n/a

It may be easier to ask yourself "Can I find discontinuities in f(x) or f'(x)?".



#3
May2406, 10:44 PM

HW Helper
P: 1,421

[tex]f(x) = \left\{ \begin{array}{ll}  x  1 & x < 1 \\ x + 1 & 1 \leq x < 0 \\ x ^ 2 + 1 & x \geq 0 \end{array} \right.[/tex] Can you follow me? Now the function is continuous on the interval [tex]( \infty , 1 )[/tex], and [tex](1, 0 )[/tex], and [tex](0 , + \infty)[/tex] right? Do you know why? Now to see if the function is continuous at 1, we simply check if: [tex]\lim_{x \rightarrow 1 ^ } f(x) = \lim_{x \rightarrow 1 ^ +} f(x) = f(1)[/tex]. If the equations above hold, then the function is continuous at x = 1. Just do the same to see if the function is continuous at x = 0. To see on which interval is he function f(x) differentiable, first you can try to find f'(x), then find where f(x) is indifferentiable (i.e where f'(x) is discontinuous or undefined). Then we can simply drop out the x values that make f(x) indifferentiable to obtain the interval on which f(x) is differentiable. Can you go from here? :) 


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